Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

PHY166 Fall 2005 7 – Impuls and Momentum; Conservation (Notions useful in the description of collisions, in particular) Newton´s second law F = ma Impuls - Momentum relation F∆t = m∆v ≡ ∆P ( a=∆v/∆t ) Impuls Definitions: Momentum conservation in isolated systems ∆t = t2−t1 etc. ( F=0 ) Change of momentum 1 – initial state ∆P = 0 P = const Momentum 2 - final state. For a system of interacting objects, adding up Newton´s second laws for all objects: F = F1+ F2 +… = m1a1+ m2a2 +… (Sum only external forces. Internal forces make no contribution according to Newton´s third law F21=−F12) F∆t = ∆P, where P = P1+ P2 +… In isolated systems, ( F=0 ) total momentum is conserved: P = const Note that P is vector! 1 Collisions and recoil A Two limiting cases of collisions: B Before (1) A B After (2) • Inelastic (objects glue together after collision; Some part of the mechanical energy is lost) PA,1+ PB,1 = P2 mAvA,1+ mBvB,1 = (mA+ mB)v2 • Elastic (Objects rebound; Mechanical energy is conserved) PA,1+ PB,1 = PA,2+ PB,2 EA,1+ EB,1 = EA,2+ EB,2 mAvA,1+ mBvB,1 = mAvA,2+ mBvB,2 mAv2A,1+ mBv2B,1 = mAv2A,2+ mBv2B,2 Conservation of momentum for inelastic collisions and conservation of momentum and energy for elastic collisions allow to solve a multitude of important problems Recoil is a phenomenon inverse to the inelastic collision: Forces acting between two objects cause them to move in different directions Example: The gun and the bullet (the gun recoils back and hits the shoulder) Whereas in the inelastic collision the energy is absorbed (mechanical energy is transformed into heat), in the recoil the energy is released (chemical energy is transformed into mechanical energy) 2 Elastic collision in one dimension (head-on collision) Transform mAvA,1+ mBvB,1 = mAvA,2+ mBvB,2 mAv2A,1+ mBv2B,1 = mAv2A,2+ mBv2B,2 m A (v A,1 − v A, 2 ) = mB (vB , 2 − vB ,1 ) as m A (v A2 ,1 − v A2 , 2 ) = mB (vB2 , 2 − vB2 ,1 ) and divide second equation over the first v A,1 + v A, 2 = vB ,1 + vB , 2 or v A,1 − vB ,1 = −(vB , 2 − v A, 2 ) (the relative velocity changes the sign after collision) For equal masses, mA= mB, the equation above can be combined with v A,1 + vB ,1 = vB , 2 + v A, 2 Leading to v A, 2 = vB ,1 , vB , 2 = v A,1 (momentum conservation) (colliding bodies exchange velocities) Demonstration 3 Center of mass (CM) of an extended system Definition: r= 1 M ∑m r i i i where M = ∑ mi total mass of a system i Acceleration of CM: Ma = ∑ mi a i = i ∑F ≡ F i - the net force (total external force) i Center of mass of an extended system moves in the same way under the influence of the net force F as the point mass M would move in this case. If we are interested in the motion of an object as a whole we can consider only the motion of the CM. If the net force is zero, CM moves with a constant velocity, whereas the object, in general, rotates around its CM. The force of gravity can be considered as applied to CM. 4 Applications, exercizes Problem A golf club exerts an average force of 500 N on a 0.1 kg golf ball, but the club is in contact with the ball for only a hundredth of a second. a) What is the magnitude of the impuls delivered by the club? b) What is the velocity acquired by the golf ball? Solution Given: F=500 N, ∆t=0.01 s, m=0.1 kg To find: a) F∆t - ? b) v - ? (this is final velocity; initial velocity is zero: v1=0, v2=v) a) F∆t = 500 N × 0.01 s = 5 N s = 5 kg m / s a) F∆t = m∆v = m(v2-v1) = mv; v = F∆t/m = 5 N s / 0.1 kg = 50 m/s Check units: N s / kg = kg m / s2 × s / kg = m/s, OK 5 Problem Four railroad cars, all with the same mass of 20000 kg sit on a track. A fifth car of identical mass approaches them with a velocity of 15 m/s. This car collides with the other four cars. a) what is the initial momentum of the system? b) what is the velocity of the five coupled cars after the collision? Solution Given: mA=4m, mB=m, m=20000 kg, vA,1=0, vB,1=15 m/s To find: a) P1 - ? b) v2 - ? a) P1 = mBvB,1 = 20000 kg 15 m/s = 300000 kg m / s b) mAvA,1+ mBvB,1 = (mA+ mB)v2, v2 = mB vB ,1 m A + mB = vA,1 = 0 mvB ,1 1 1 = vB ,1 = 15 m/s = 3 m/s 5m 5 5 6 Problem Calculate the fraction of the energy lost in the railroad car collision of the problem above Solution 2 2 (m A + mB )v22 mvB ,1 5m(vB ,1 / 5) Elost = E1 − E2 = − = − 2 2 2 2 mvB2 ,1 1 4 mvB2 ,1 4 = = E1 1 − = 2 5 5 2 5 mB vB2 ,1 Fraction of the energy lost: Elost 4 = , that is, 80% E1 5 7 Problem Billiard ball A moving with a speed 3 m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45° to the x axis, ball A above the x axis and ball B below. What are the speeds of the two ball after the collision? Solution Given: mA=mB, vA,1=3 m/s, vB,1= 0, θA,1=45°, θB,1=−45° To find: v2,A - ? v2,B - ? Momentum of the isolated system of two balls is conserved: P1 = P1, A + P1, B = P2 = P2, A + P2, B Momentum conservation along the y axis: P1, y = 0 = P2, y = m(v2, A, y + v2, B , y ) = m(v2, A sin(45°) + v2, B sin(−45°) ) = m ( v2 , A − v2 , B ) 2 v 2 , A = v2 , B Momentum conservation along the x axis: P1, x ≡ mv1, A = P2, x = m(v2, A, x + v2, B , x ) = m(v2, A cos(45°) + v2, B cos(−45°) ) = m (v2, A + v2, B ) = 2mv2, A 2 v2 , A = v2 , B = 8 = 2.1 m/s v1, A 2 Homework: Giancoli Chapter 7, Problems 7, 17, 23, 29, 51 9